Suppose $O$ is an open subset of $\mathbb{R}^n$ and let $\phi\colon O\to \mathbb{R}^n$ be a $C^1$ function. Suppose $\alpha\in O$ and consider the derivative $D\phi(\alpha)$ at $\alpha$.
I am trying to show that if $D\phi(\alpha)$ is invertible then there is a neighborhood $V$ of $\alpha$ and a $\delta>0$ such that for all $x,y\in U$, $\|f(x)-f(y)\|\geq \delta\|x-y\|$.
I am pretty much stuck on how to proceed, so any help will be appreciated. Thanks.
Since $D\phi(\alpha)$ is invertible, there is a neighborhood $V$ of $\alpha$ such that for all $u\in V$ the matrix $D\phi(u)$ is invertible and (a) holds. Now, for $u\in V$, let $g(u)=D\phi(u)^{-1}$ be the inverse of $D\phi(u)$. Then $g(u)$ is continuous in $u$ and so there is a neighborhood $U$ of $\alpha$ such that (b) holds. Let $\delta=\frac{\varepsilon}{\|g(\alpha)\|}$ where $\varepsilon>0$ is such that (c) holds. Now, suppose $x,y\in U$. Then, $$\|f(x)-f(y)\|=\|D\phi(x)g(x)-D\phi(y)g(y)\|\leq \|D\phi(x)\|\|g(x)-g(y)\|+\|D\phi(x)g(y)-D\phi(y)g(y)\|.$$ Now, for $i=1,\ldots,n$, we have $$\|D\phi(x)g(x)-D\phi(y)g(y)\|_i\leq \sum_{j=1}^n\left|\frac{\partial\phi_i}{\partial x_j}(x)-\frac{\partial\phi_i}{\partial x_j}(y)\right||g(x)-g(y)|_j$$ and $$\|D\phi(x)g(y)-D\phi(y)g(y)\|_i\leq \sum_{j=1}^n\left|\frac{\partial\phi_i}{\partial x_j}(x)-\frac{\partial\phi_i}{\partial x_j}(y)\right||g(y)|_j.$$ Since $g(u)$ is continuous in $u$, $\|g(x)-g(y)\|\leq \|g(\alpha)\|\|x-y\|$. Also, since $D\phi(u)$ is continuous in $u$, $\|D\phi(x)-D\phi(y)\|\leq \|D\phi(\alpha)\|\|x-y\|$. Therefore, $$\|f(x)-f(y)\|\leq n\|D\phi(\alpha)\|\|g(\alpha)\|\|x-y\|.$$ Now, since $\|g(\alpha)\|$ and $\|D\phi(\alpha)\|$ are both bounded above by a constant, we have $$\|f(x)-f(y)\|\leq nM\|x-y\|$$ for some constant $M$ (in fact, $M=\|g(\alpha)\|\|D\phi(\alpha)\|$). Therefore, if $x,y\in U$, then $$\|f(x)-f(y)\|\geq \|x-y\|\frac{\varepsilon}{nM}=\delta\|x-y\|.$$